Step 1: Given condition
We are given $\sin \theta = \cos \theta$.
Step 2: Divide both sides by $\cos \theta$ (valid for $0^\circ \leq \theta \leq 90^\circ$, except $\theta=90^\circ$)
\[ \frac{\sin \theta}{\cos \theta} = \frac{\cos \theta}{\cos \theta} \] \[ \tan \theta = 1 \]
Step 3: Solve for $\theta$
\[ \tan \theta = 1 \,\Rightarrow\ \theta = 45^\circ \]
Step 4: Check interval
Since $0^\circ \leq \theta \leq 90^\circ$, the only solution is $\theta = 45^\circ$.
\[ \boxed{\theta = 45^\circ} \]
The product of $\sqrt{2}$ and $(2-\sqrt{2})$ will be:
If a tangent $PQ$ at a point $P$ of a circle of radius $5 \,\text{cm}$ meets a line through the centre $O$ at a point $Q$ so that $OQ = 12 \,\text{cm}$, then length of $PQ$ will be:
In the figure $DE \parallel BC$. If $AD = 3\,\text{cm}$, $DE = 4\,\text{cm}$ and $DB = 1.5\,\text{cm}$, then the measure of $BC$ will be: